Question

When preparing the following solutions, indicate each concentration by mass percentage (%), mass to volume percentage (w/v%) and molar concentration (M). WWW (1) Preparation of diluted ammonia water: Add pure water to 400 mL of strong ammonia water (28%, specific weight: 0.9, molecular weight: 17.0) to make 1000 mL. (2) Preparation of dilute hydrochloric acid: Add pure water to 23.6 mL of hydrochloric acid (36.5%, specific weight: 1.18, molecular weight: 36.5) to make 100 mL. (3) Dilute nitric acid: Add pure water to 10.5 mL of nitric acid (70%, specific weight: 1.42, molecular weight 63.0) to make 100 mL. (4) Dilute sulfuric acid: Add 5.7 mL of sulfuric acid (98%, specific weight: 1.84, molecular weight: 98.0) to 10 mL of pure water, cool with stirring, and add pure water to make 100 mL. www

Answer 1

(1) Given: original solution of ammonia water is 28%
Initial volume = 400 mL
Density/specific weight = 0.9 g/mL
Mass of solution = Density x volume = 0.9 g/mL x 400 mL = 360 g
Mass od NH₃ in the sample = 28% of 360 g = 0.28 x 360 = 100.8 g
600 mL of pure water is added to make the final volume = 1000 mL
m/v % = (100.8 g/1000 mL) x 100 = 10.08 %
Density of water = 1.0 g /mL
Hence mass of the 600 mL of water added = 600 g
Total mass of the solution = 360 + 600 = 960 g
m/m% = (100.8 g/960 g) x 100 = 10.5 %
Number of moles of NH₃ = Given mass/Molar mass = 100.8 g/17.0 g = 5.929
Volume of solution = 1000 mL = 1 L
Molarity = Number of moles /Vol. in L = 5.929 M

2. Given: original solution of HCl is 36.5 %
Initial volume = 23.6 mL
Density/specific weight = 1.18 g/mL
Mass of solution = Density x volume = 1.18 g/mL x 23.6 mL = 27.848 g
Mass of HCl in the sample = 36.5% of 27.848 g = 0.365 x 27.848 = 10.16452 g
76.4 mL of pure water is added to make the final volume = 100 mL
m/v % = (10.16452 g/100 mL) x 100 = 10.16452 %
Density of water = 1.0 g /mL
Hence mass of the 76.4 mL of water added = 76.4 g
Total mass of the solution = 27.848+ 76.4 = 104.248 g
m/m% = (10.16452 g/104.248 ​​​​​​​g) x 100 = 9.75 %
Number of moles of HCl = Given mass/Molar mass = 10.16452 g/36.5 g = 0.27848
Volume of solution = 100 mL = 0.100 L
Molarity = Number of moles /Vol. in L = 0.27848/0.100 L = 2.7848 M